In the decay sequence, \[_{92}^{238}U\xrightarrow{-{{X}_{1}}}_{90}^{234}Th\xrightarrow{-{{X}_{2}}}_{91}^{234}Pa{{\xrightarrow{-{{X}_{3}}}}^{234}}Z\xrightarrow{-{{X}_{4}}}_{90}^{234}Th\], ${{X}_{1}}$,${{X}_{2}}$,${{X}_{3}}$ and ${{X}_{4}}$ are particle/radiation emitted by the respective isotopes . The correct option(s) is (are)
This question has multiple correct options
(A) ${{X}_{1}}$will deflect towards negatively charged plate
(B) ${{X}_{2}}$is ${{\beta }^{-}}$
(C) ${{X}_{3}}$is γ- ray
(D) z is an isotope of uranium

Hint: The spontaneous breakdown of an atomic nucleus resulting in the release of matter and energy from the nucleus can be termed as the radioactive decay. Mainly there are five types of radioactive decay such as alpha emission, beta emission, gamma emission, electron capture, and positron emission and each type of these decay emits a specific particle which changes upon the type of product produced.

Complete step by step solution:
- Radioactive decay can be defined as the loss of elementary particles from an unstable nucleus which ultimately changes an unstable element into another more stable element.
- An alpha particle (α) is composed of two protons and two neutrons and in Alpha Decay, an alpha particle is emitted from the nucleus and in Beta Decay (β), a neutron from an atom will split into a negatively charged electron and one positively charged proton. Also, the Gamma rays are electromagnetic waves with very high frequencies and energy in Gamma Decay (γ), energy in the form of gamma rays or radiation are emitted from the nucleus.
- The first step in the given radioactive decay can be written as follows,
\[_{92}^{238}U\xrightarrow{{}}_{90}^{234}Th+_{2}^{4}He\]
As we can see an alpha particle (α) is being emitted and hence ${{X}_{1}}$ is an α particle and it will deflect towards the negatively charged plate. Thus option (A) is correct.
 - The second step in the given radioactive decay can be written as follows
\[_{90}^{234}Th\xrightarrow{{}}_{91}^{234}Pa+_{-1}^{0}e\]
As we can observe a beta particle (β) is being emitted and hence ${{X}_{2}}$ is ${{\beta }^{-}}$.Thus option (B) is also correct.
-The third step of the radioactive decay can be written as follows
\[_{91}^{234}Pa{{\xrightarrow{{}}}^{234}}Z+_{-1}^{0}e\]
Here also a beta particle (β) is being emitted and hence is${{X}_{3}}$ is ${{\beta }^{-}}$ not γ- ray. Hence option (C) is incorrect. Also $^{234}Z.$ is an isotope of uranium since it has an atomic number of 92 which is the same as the atomic number of uranium. Hence option (D) is also correct.

Therefore correct options are (A), (B) and (D).

Note:
Bear in mind that, the radioactive decay can also be expressed in terms of entropy, as the tendency for matter and energy to gain inert stability or uniformity. Also, for a given element, the disintegration or decay rate will be proportional to the number of atoms and the activity I usually measured in terms of atoms per unit time.